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I am not 100% sure if this statement is even true, but I think it is. Any help would be great, I have tried to prove this, but I really haven't gotten anywhere constructive with this,

If $a+b = 1$ and $0 < a, b < 1$, prove that $ 4 \leq \frac{1}{a} + \frac{1}{b}$.

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up vote 2 down vote accepted

Recall the AM-GM-HM inequality. If $a_1,a_2,\ldots,a_n$ are $n$ positive numbers, then $$\underbrace{\dfrac1n \left(\displaystyle \sum_{k=1}^n a_k\right)}_{AM} \geq \underbrace{\sqrt[n]{a_1a_2\cdots a_n}}_{GM} \geq \underbrace{\left(\dfrac1n \displaystyle \sum_{k=1}^n a_k^{-1}\right)^{-1}}_{HM}$$ In your case, $n=2$ and comparing the AM and HM, we get that $$\dfrac{a+b}2 \geq \left(\dfrac12 \left(\dfrac1a + \dfrac1b\right)\right)^{-1}$$

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I find these proofs pleasing. +1 – Rustyn Jan 27 '13 at 6:03

$$a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0$$ so that $$ \sqrt{ab} \leq \frac{a+b}{2} = \frac{1}{2} \Rightarrow ab \leq \frac{1}{4}$$

$$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{1}{ab}\geq \frac{1}{1/4}=4$$

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While not giving away the solution, here are some pretty clear (hopefully) steps:

  1. Try simplifying the LHS of the inequality.
  2. Make use of the fact that you're given $a+b=1$
  3. Solve for the only variable left
  4. At this point, recall what you did in step 1, then make use of the variable you solved for in step 3 (the product of $a$ and $b$).

You should then arrive at your solution.

It's just as Andre Nicolas mentions.

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All you need to show is that if $a+b=1$ (and $a$ and $b$ are positive) then $ab\le \frac{1}{4}$.

This can be done in various ways. For example, by AM-GM, $\frac{a+b}{2}\ge \sqrt{ab}$.

Or else use elementary calculus: maximize $a(1-a)$, given that $0\lt a\lt 1$, though that restriction is not really relevant. Or else we can maximize by completing the square.

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